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# Posts Tagged ‘average occupancy’

## A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

## Fermi-Dirac function expansion for thermodynamic quantities

Posted by peeterjoot on April 21, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

In section 8.1 of [1] are some Fermi-Dirac \index{Fermi-Dirac function} expansions for $P$, $N$, and $U$. Let’s work through these in detail.

Our starting point is the relations

\begin{aligned}P V \beta = \ln Z_{\mathrm{G}} = \sum \ln \left( 1 + z e^{-\beta \epsilon} \right)\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}N = \sum \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }}.\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Recap. Density of states

We’ll employ the 3D non-relativisitic density of states

\begin{aligned}\mathcal{D}(\epsilon) &= \sum_\mathbf{k} \delta(\epsilon - \epsilon_\mathbf{k}) \\ &\sim V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\delta(\epsilon - \epsilon_\mathbf{k}) \\ &= \frac{4 \pi V}{(2 \pi)^3}\int dk k^2 \delta\left( \epsilon - \frac{\hbar^2 k^2}{2 m} \right) \\ &= \frac{4 \pi V}{(2 \pi)^3}\int dk k^2 \frac{ \delta\left( k - \sqrt{2 m \epsilon}/\hbar \right)}{ \frac{\hbar^2}{m} \frac{\sqrt{2 m \epsilon}}{\hbar}} \\ &= \frac{2 V}{(2 \pi)^2 }\frac{m}{\hbar^2}\sqrt{\frac{2 m \epsilon}{\hbar^2}},\end{aligned} \hspace{\stretch{1}}(1.0.1b)

or

\begin{aligned}\boxed{\mathcal{D}(\epsilon)=\frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\epsilon^{1/2}.}\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Density

Now let’s make our integral approximation of the sum for $N$. That is

\begin{aligned}N &= g \int d\epsilon \mathcal{D}(\epsilon) \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} \\ &= g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \frac{\epsilon^{1/2}}{ z^{-1} e^{\beta \epsilon} + 1 } \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty du \frac{u^{1/2}}{ z^{-1} e^{u} + 1 } \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\left( \frac{2 m}{\hbar^2} \right)^{3/2}\Gamma(3/2) f_{3/2}(z) \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{\hbar^3}\frac{1}{{2}} \sqrt{\pi}f_{3/2}(z)\\ &= g V \not{{2}} \pi\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3}\frac{1}{{\not{{2}}}} \sqrt{\pi}f_{3/2}(z),\end{aligned} \hspace{\stretch{1}}(1.0.1b)

or

\begin{aligned}\frac{N}{V} = g \frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3}f_{3/2}(z).\end{aligned} \hspace{\stretch{1}}(1.0.5)

With

\begin{aligned}\lambda = \frac{h}{\sqrt{ 2 \pi m k_{\mathrm{B}} T }},\end{aligned} \hspace{\stretch{1}}(1.0.6)

this gives us the desired density result from the text

\begin{aligned}\boxed{\frac{N}{V}=\frac{g}{\lambda^3} f_{3/2}(z).}\end{aligned} \hspace{\stretch{1}}(1.0.7)

Pressure

For the pressure, we can do the same, but have to integrate by parts

\begin{aligned}P V \beta &= g \sum \ln \left( 1 + z e^{-\beta \epsilon} \right) \\ &\sim g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \epsilon^{1/2} \ln \left( 1 + z e^{-\beta \epsilon} \right) \\ &= - g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \frac{2}{3} \epsilon^{3/2} \frac{-\beta z e^{-\beta \epsilon} }{ 1 + z e^{-\beta \epsilon} } \\ &= g\frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\frac{2}{3} \frac{1}{{\beta^{3/2}}}\int_0^\infty dx\frac{x^{3/2}}{z^{-1} e^{x} + 1 } \\ &= g\frac{2}{3} 2 \pi V\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }\Gamma(5/2)f_{5/2}(z) \\ &= g\frac{2}{3} 2 \pi V\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }\frac{3}{2} \frac{1}{2} \sqrt{\pi}f_{5/2}(z) \\ &= g V\frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }f_{5/2}(z),\end{aligned} \hspace{\stretch{1}}(1.0.7)

or

\begin{aligned}\boxed{P \beta = \frac{g}{\lambda^3} f_{5/2}(z).}\end{aligned} \hspace{\stretch{1}}(1.0.9)

Energy

The average energy is the last thermodynamic quantity to come very easily. We have

\begin{aligned}U &= - \frac{\partial {}}{\partial {\beta}} \ln Z_{\mathrm{G}} \\ &= - \frac{\partial {T}}{\partial {\beta}} \frac{\partial {}}{\partial {T}} \ln Z_{\mathrm{G}} \\ &= - \frac{\partial {(1/k_{\mathrm{B}} T)}}{\partial {\beta}} \frac{\partial {}}{\partial {T}} P V \beta \\ &= \frac{1}{{k_{\mathrm{B}} \beta^2}}\frac{\partial {}}{\partial {T}} \frac{g V}{\lambda^3} f_{5/2}(z) \\ &= g V k_{\mathrm{B}} T^2f_{5/2}(z)\frac{\partial {}}{\partial {T}} \frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3} \\ &= \frac{3}{2} \frac{g V k_{\mathrm{B}} T}{\lambda^3}f_{5/2}(z).\end{aligned} \hspace{\stretch{1}}(1.0.9)

From eq. 1.0.7, we have

\begin{aligned}\frac{g V}{\lambda^3} = \frac{N}{f_{3/2}(z) },\end{aligned} \hspace{\stretch{1}}(1.0.11)

so the energy takes the form

\begin{aligned}\boxed{U = \frac{3}{2} N k_{\mathrm{B}} T \frac{f_{5/2}(z)}{f_{3/2}(z) }.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

We can compare this to the ratio of pressure to density

\begin{aligned}\frac{P \beta}{n} = \frac{f_{5/2}(z)}{f_{3/2}(z) },\end{aligned} \hspace{\stretch{1}}(1.0.11)

to find

\begin{aligned}U= \frac{3}{2} N k_{\mathrm{B}} T \frac{P V \beta}{N}= \frac{3}{2} P V,\end{aligned} \hspace{\stretch{1}}(1.0.11)

or

\begin{aligned}\boxed{P V = \frac{2}{3} U.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## Pathria chapter 4 diatomic molecule problem

Posted by peeterjoot on March 18, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Diatomic molecule ([1] pr 4.7)

Consider a classical system of non-interacting, diatomic molecules enclosed in a box of volume $V$ at temperature $T$. The Hamiltonian of a single molecule is given by

\begin{aligned}H(\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2) = \frac{1}{{2m}} \left( \mathbf{p}_1^2 + \mathbf{p}_2^2 \right)+\frac{1}{{2}} K \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Study the thermodynamics of this system, including the dependence of the quantity $\left\langle{{r_{12}^2}}\right\rangle$ on $T$.

Partition function
First consider the partition function for a single diatomic pair

\begin{aligned}Z_1 &= \frac{1}{{h^6}} \int d^6 \mathbf{p} d^6 \mathbf{r} e^{-\beta \frac{ \mathbf{p}_1^2 + \mathbf{p}_2^2 }{2m}} e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}} \\ &= \frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{6/2}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Now we can make a change of variables to simplify the exponential. Let’s write

\begin{aligned}\mathbf{u} = \mathbf{r}_1 - \mathbf{r}_2\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{v} = \mathbf{r}_2,\end{aligned} \hspace{\stretch{1}}(1.0.3b)

or

\begin{aligned}\mathbf{r}_2 = \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\mathbf{r}_1=\mathbf{u} + \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Our volume element is

\begin{aligned}d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 = d^3 \mathbf{u} d^3 \mathbf{v} \frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

It wasn’t obvious to me that this change of variables preserves the volume element, but a quick Jacobian calculation shows this to be the case

\begin{aligned}\frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})} &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ \partial r_{21}/\partial u_1 & \partial r_{21}/\partial u_2 &\partial r_{21}/\partial u_3 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ \partial r_{22}/\partial u_1 & \partial r_{22}/\partial u_2 &\partial r_{22}/\partial u_3 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ \partial r_{23}/\partial u_1 & \partial r_{23}/\partial u_2 &\partial r_{23}/\partial u_3 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= 1.\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our remaining integral can now be evaluated

\begin{aligned}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}} &= \int d^3 \mathbf{u} d^3 \mathbf{v} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \left( \frac{ 2 \pi }{ K \beta } \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Our partition function is now completely evaluated

\begin{aligned}Z_1 = V\frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{3}\left( \frac{ 2 \pi }{ K \beta } \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

As a function of $V$ and $T$ as in the text, we write

\begin{aligned}Z_1 = V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}f(T) = \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

Gibbs sum

Our Gibbs sum, summing over the number of molecules (not atoms), is

\begin{aligned}Z_{\mathrm{G}} &= \sum_{N_r = 0}^\infty \frac{z^{N_r}}{N_r!} Z_1^{N_r} \\ &= e^{ z V f(T) },\end{aligned} \hspace{\stretch{1}}(1.0.10)

or

\begin{aligned}q &= \ln Z_{\mathrm{G}} \\ &= z V f(T) \\ &= P V \beta.\end{aligned} \hspace{\stretch{1}}(1.0.11)

The fact that we can sum this as an exponential series so nicely looks like it’s one of the main advantages to this grand partition function (Gibbs sum). We can avoid any of the large $N!$ approximations that we have to use when the number of particles is explicitly fixed.

Pressure

The pressure follows

\begin{aligned}P &= z f(T) k_{\mathrm{B}} T \\ &= e^{\mu/k_{\mathrm{B}} T}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{11/2}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Average energy

\begin{aligned}\left\langle{{H}}\right\rangle &= -\frac{\partial {q}}{\partial {\beta}} \\ &= - z V \frac{9}{2} \frac{f(T)}{T} \frac{\partial {T}}{\partial {\beta}} \\ &= z V \frac{9}{2} \frac{f(T)}{T^3} \frac{1}{{k_{\mathrm{B}}}},\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}\left\langle{{H}}\right\rangle = e^{\mu/k_{\mathrm{B}} T} V \frac{9}{2} k_{\mathrm{B}}^2 \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Average occupancy

\begin{aligned}\left\langle{{N}}\right\rangle &= z \frac{\partial {}}{\partial {z}} \ln Z_{\mathrm{G}} \\ &= z \frac{\partial {}}{\partial {z}} \left( z V f(T) \right) \\ &= z V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.15)

but this is just $q$, or

\begin{aligned}\left\langle{{N}}\right\rangle &= e^{\mu/k_{\mathrm{B}} T} V\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Free energy

\begin{aligned}F &= - k_{\mathrm{B}} T \ln \frac{ Z_{\mathrm{G}} }{z^N} \\ &= - k_{\mathrm{B}} T \left( q - N \ln z \right) \\ &= N k_{\mathrm{B}} T \beta \mu - k_{\mathrm{B}} T q \\ &= z V f(T) \mu - k_{\mathrm{B}} T z V f(T) \\ &= z V f(T) \left( \mu - k_{\mathrm{B}} T \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

\begin{aligned}F = e^{\mu/k_{\mathrm{B}} T} V \left( \mu - k_{\mathrm{B}} T \right)\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

Entropy

\begin{aligned}S &= \frac{U - F}{T} \\ &= \frac{V}{T} e^{\mu/k_{\mathrm{B}} T} \left( k_{\mathrm{B}} T \right)^{3/2}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}} \right)^3\left( \frac{9}{2} k_{\mathrm{B}}^2 - \left( \mu - k_{\mathrm{B}} T \right) \left( k_{\mathrm{B}} T \right)^3 \right).\end{aligned} \hspace{\stretch{1}}(1.0.19)

Expectation of atomic separation

The momentum portions of the average will just cancel out, leaving just

\begin{aligned}\left\langle{r_{12}^2}\right\rangle &= \frac{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }}{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }} \\ &= \frac{ \int d^3 \mathbf{u} \mathbf{u}^2 e^{-\beta K \mathbf{u}^2 /2 }}{\int d^3 \mathbf{u} e^{-\beta K \mathbf{u}^2 /2 }} \\ &= \frac{\int da db dc \left( a^2 + b^2 + c^2 \right) e^{-\beta K \left( a^2 + b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2}} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}\int db dc e^{-\beta K \left( b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2 }} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}}{\int e^{-\beta K a^2/2}}\end{aligned} \hspace{\stretch{1}}(1.0.20)

Expanding the numerator by parts we have

\begin{aligned}\int da a^2 e^{-\beta K a^2/2} \\ &= \int a d\frac{ e^{-\beta K a^2/2}}{- 2 \beta K/2} \\ &= \frac{1}{\beta K}\int e^{-\beta K a^2/2}.\end{aligned} \hspace{\stretch{1}}(1.0.21)

This gives us

\begin{aligned}\boxed{\left\langle r_{12}^2 \right\rangle = \frac{3}{\beta K} = \frac{3 k_{\mathrm{B}} T}{K}.}\end{aligned} \hspace{\stretch{1}}(1.0.22)

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.